Understanding Tensors and Their Role in General Relativity

[AdrianZ]

Hi all,
first of all I'm sorry if I've posted my question on the wrong section, didn't know where I should post it, I thought this forum might be the right choice.
I've recently graduated from high school and I'm highly interested in stuff about General relativity and Quantum mechanics and geometry, I have read linear algebra and except Eigenvalues I'm good at it, I also know some random stuff from Calculus and differential geometry like total differential, directional derivative, chain rule, Frenet frame and so forth. can i start reading about tensors?
what do i have to do to get enough required knowledge to understand tensors, manifolds and things required to read the GR and fluid mechanics?

I know my questions might be so general. so forgive my ignorance. any kind of help would be appreciated.

 

[dx]

Linear algebra is enough to learn about tensors. Do you know about vector spaces and linear transformations?

 

[AdrianZ]

Linear algebra is enough to learn about tensors. Do you know about vector spaces and linear transformations?

Yes I do. I have no problem with linear algebra except where it comes to eigenvalues and eigenvectors. another thing that I don't understand well is Gram-schmidt process. but vector spaces and linear transformations are easy to understand for me.
won't I need to know some calculus? I have watched a series of Susskind's general relativity lectures on youtube and he talked about some calculus topics I think. would you please explain the prerequisites I need to know for tensors?

 

[dx]

Tensors are very simple. They are simply linear functions, i.e. functions which are linear in each of their arguments; e.g. f(a + b) = f(a) + f(b). So to understand their basic nature, you don't need to know a lot. If you have a vector space V, then the linear functions f : V → R are called covectors, the simplest type of tensor. The space of covectors on V is also a vector space, called V*. A general tensor is a linear function from m copies of V and n copies of V*, i.e. it is a function with m + n arguments, where the first m are from V and the rest are from V*.

If you want to learn general relativity, you have to learn about manifolds. There can be tensor fields on manifolds, and the description of how they change and vary requires calculus.

 

[adriank]

Linear algebra is not enough to learn about tensor fields, which is what physicists mean when they talk about tensors. That stuff is handled by differential geometry; in particular, (pseudo-)Riemannian geometry for GR.

 

[AdrianZ]

Tensors are very simple. They are simply linear functions, i.e. functions which are linear in each of their arguments; e.g. f(a + b) = f(a) + f(b). So to understand their basic nature, you don't need to know a lot. If you have a vector space V, then the linear functions f : V → R are called covectors, the simplest type of tensor. The space of covectors on V is also a vector space, called V*. A general tensor is a linear function from m copies of V and n copies of V*, i.e. it is a function with m + n arguments, where the first m are from V and the rest are from V*.

If you want to learn general relativity, you have to learn about manifolds. There can be tensor fields on manifolds, and the description of how they change and vary requires calculus.

Thank you, your help is really useful.
let me see if I've understood you correctly. can we say that technically a tensor is an object in multi-linear algebra? I mean you said tensors are linear functions in each of their arguments. can we say determinants are a type of tensors because they are linear in each of their columns and rows?
you said a covector is a function f: V -> R. R here denotes the Real line? right? so if f is a function with three inputs x,y,z where x,y,z are vectors, then A[x,y,z] which is the determinant made up of these 3 vectors as its columns is a covector? and then all determinants of such 3 vectors form a vector space because we can add them and multiply them by scalars? am I lost or I've understood it? sorry If my questions are wrong,I'm a beginner.

what things do I need to know to understand manifolds?

 

[AdrianZ]

Linear algebra is not enough to learn about tensor fields, which is what physicists mean when they talk about tensors. That stuff is handled by differential geometry; in particular, (pseudo-)Riemannian geometry for GR.

I know some simple stuff from differential geometry. like curvature of a curve, torsion of a curve, vector fields, covariant derivative, exterior derivative and some other simple stuff. what things do I have to read from differential geometry? is there a way to learn tensors intuitively with geometrical properties instead of learning them like the way they teach vectors in vector analysis?

 

[dx]

can we say that technically a tensor is an object in multi-linear algebra?

Yes, a tensor is essentially a multilinear function.

can we say determinants are a type of tensors because they are linear in each of their columns and rows?

Yes, the determinant is a tensor.

you said a covector is a function f: V -> R. R here denotes the Real line? right?

Yes

so if f is a function with three inputs x,y,z where x,y,z are vectors, then A[x,y,z] which is the determinant made up of these 3 vectors as its columns is a covector?

No, a covector is defined with respect to some vector space V, and has only one argument. A tensor with three arguments would be called a 0-3 tensor, or a 3-0 tensor, or a 2-1 tensor etc. depending on how many vector and covector arguments it has.

what things do I need to know to understand manifolds?

Calculus and linear algebra are the essentials.

 

[AdrianZ]

No, a covector is defined with respect to some vector space V, and has only one argument. A tensor with three arguments would be called a 0-3 tensor, or a 3-0 tensor, or a 2-1 tensor etc. depending on how many vector and covector arguments it has.

Calculus and linear algebra are the essentials.

well, let me try again. for example the Euclidean length of a vector is a covector that we can define it like l(x) = $$\sqrt{x.x}$$? right? if yes, you said that the set of covectors defined with respect to some vector space V form a new vector space called V*. so if we take another covector like f(x) = 5x.x then all linear combinations like af(x) + bl(x) are also covectors and they create a new vector space called V*?
would you give me an example of how to define a tensor with the definition you gave me?

do I need to know differential equations as well? the book I'm reading for Calculus is by Tom M. Apostol, it talks a lot about differential equations in it. It's way harder than Barret O'neills differential geometry that I'm reading. can I read O'neills' diff geometry instead?

 

[arkajad]

Strictly speaking: determinant is usually, when it appear is differential-geometric context, a tensor density.

Tensors are geometric objects associated to the tangent space at a given point.

Tensor fields are tensors (usually smoothly) depending on the point.

In order to study tensors - all you need is multilinear algebra.

To study tensor fields, and natural operations on these fields - you need differential geometry.

 

[dx]

for example the Euclidean length of a vector is a covector that we can define it like l(x) = $$\sqrt{x.x}$$?

The length function is not linear, because the length of u + v is not the sum of the length of u and the length of v.

would you give me an example of how to define a tensor with the definition you gave me?

Here's a simple example: let's consider the vector space R², i.e. the space of ordered pairs of real numbers {(t, x)| t & x in R}. The function which takes (t, x) to t is linear, because (t + t', x + x') wold go to t + t'. This function, which is a covector is a simple version of what in differential geometry is called dt. The function which takes (t, x) to x would be dx.

Now using these two covectors, i.e. dx and dt, we can form another linear function of two arguments, whose result is the product of the two original functions on the two arguments. This tensor is called $$dx \otimes dt$$. We can even add tensors to get new tensors.

In special relativity, we have a vector space called Minkowski space, which represents space and time. From the coordinates t, x, y, z of spacetime, we can obtain the covectors dt, dx, dy and dz. The Minkowski metric tensor can then be expressed as

$$dt \otimes dt - dx \otimes dx - dy \otimes dy - dz \otimes dz$$

do I need to know differential equations as well? the book I'm reading for Calculus is by Tom M. Apostol. It's way harder than Barret O'neills differential geometry that I'm reading. can I read O'neills' diff geometry instead?

You don't need a great deal of knowledge about differential equations to understand of manifolds and differential geometry, but you should know calculus well enough that the notion of 'differential equation' is completely clear to you.

 

[AdrianZ]

Here's a simple example: let's consider the vector space R², i.e. the space of ordered pairs of real numbers {(t, x)| t & x in R}. The function which takes (t, x) to t is linear, because (t + t', x + x') wold go to t + t'. This function, which is a covector is a simple version of what in differential geometry is called dt. The function which takes (t, x) to x would be dx.

Now using these two covectors, i.e. dx and dt, we can form another linear function of two arguments, whose result is the product of the two original functions on the two arguments. This tensor is called $$dx \otimes dt$$. We can even add tensors to get new tensors.

In special relativity, we have a vector space called Minkowski space, which represents space and time. From the coordinates t, x, y, z of spacetime, we can obtain the covectors dt, dx, dy and dz. The Minkowski metric tensor can then be expressed as

$$dt \otimes dt - dx \otimes dx - dy \otimes dy - dz \otimes dz$$

okay, thank you.
you said:

Now using these two covectors, i.e. dx and dt, we can form another linear function of two arguments, whose result is the product of the two original functions on the two arguments.

you didn't specify what type of product you're talking about. that is a little bit confusing for me. for example can differential forms with the operation 'exterior derivative' be used to form tensors?

and my last question, do you suggest any good book that I can use to understand differential geometry for the GR?

 

[AdrianZ]

Strictly speaking: determinant is usually, when it appear is differential-geometric context, a tensor density.

Tensors are geometric objects associated to the tangent space at a given point.

Tensor fields are tensors (usually smoothly) depending on the point.

In order to study tensors - all you need is multilinear algebra.

To study tensor fields, and natural operations on these fields - you need differential geometry.

thank you

 

[dx]

you didn't specify what type of product you're talking about. that is a little bit confusing for me. for example can differential forms with the operation 'exterior derivative' be used to form tensors?

Say we have two covectors, A : V → R and B : V → R. Then the action of $A \otimes B$ on (u, v) is A(u)A(v), which is a product of real numbers. ($A \otimes B$ is called the tensor product of A and B.)

About exterior derivatives: yes the exterior derivative acts on differential forms to give differential forms. Differential forms are antisymmetric tensors. The determinant is an example of a differential form because it is antisymmetric: D(u,v) = -D(v,u).

and my last question, do you suggest any good book that I can use to understand differential geometry for the GR?

My favorite book on this topic, at least for basics, is "Spacetime, Geometry and Cosmology" by William Burke.

 

[arkajad]

I would also suggest a really comprehensive one: "Differential Geometry and Lie Groups for Physicists" by Marian Fecko. If you can - grab it.

 

[arkajad]

The determinant is an example of a differential form because it is antisymmetric: D(u,v) = -D(v,u).

In GR we often have to deal with $$\sqrt{|\det (g)|}$$

It is not exactly a differential form. It is a tensor density. Yes, it is a section of a GL(n) associated vector bundle, but it is not a differential form because of its sign, which is an invariant.

 

[AdrianZ]

Say we have two covectors, A : V → R and B : V → R. Then the action of $A \otimes B$ on (u, v) is A(u)A(v), which is a product of real numbers. ($A \otimes B$ is called the tensor product of A and B.)

About exterior derivatives: yes the exterior derivative acts on differential forms to give differential forms. Differential forms are antisymmetric tensors. The determinant is an example of a differential form because it is antisymmetric: D(u,v) = -D(v,u).

My favorite book on this topic, at least for basics, is "Spacetime, Geometry and Cosmology" by William Burke.

well, I remember that I have already read about tensor product,tensor product is just like the way we multiply polynomials, am i right? for example if tensor product acts on two vectors in 3-D space it will give us a 9-D vector which is a tensor of rank-2 and can be represented as a matrix. it seems that your definition is a reduced form of that product because you were working in R one. am i right?

now let me guess, in differential geometry they want to see, for example, how fast a tensor changes with respect to time? or how it changes with respect to displacement and things like that? in fact what I'm trying to ask is, do we have the same operations we do with vectors for tensors? and why do we need tensors? we can intuitively understand vectors and use them, can we do the same with tensors? I meant is there any intuition behind tensors and where they've come from? I guess the answer would be no becuse they are in dimensions greater than 3. am i right?

 

[dx]

In GR we often have to deal with $$\sqrt{|\det (g)|}$$

It is not exactly a differential form. It is a tensor density. Yes, it is a section of a GL(n) associated vector bundle, but it is not a differential form because of its sign, which is an invariant.

You're probably right, I don't know what a tensor density is. When I said the determinant is a differential form, I was thinking of the oriented area spanned by two vectors u and v. This is a 2-form because of its linearity and antisymmetry.

 

[dx]

in fact what I'm trying to ask is, do we have the same operations we do with vectors for tensors? and why do we need tensors? we can intuitively understand vectors and use them, can we do the same with tensors?

Tensors are more abstract than vectors, but you can build an intuition for them by considering examples. Calculus is about linear approximations to objects, and for this reason tensors are very important in the modern form of calculus, called 'calculus on manifolds'. They also have geometric representations, so if you like thinking visually, you can build intuition for them by thinking using those representations. The book I mentioned before covers these things.

 

[arkajad]

You're probably right, I don't know what a tensor density is.

Tensor density of weight w has an additional factor in the transformation law which is

$$|\det{\frac{\partial x}{\partial x'}}|^w$$

 

[AdrianZ]

I would also suggest a really comprehensive one: "Differential Geometry and Lie Groups for Physicists" by Marian Fecko. If you can - grab it.

thank you. I will look for it. I hope I would have time to read both of them soon.

 

[Rasalhague]

Tensors are more abstract than vectors

Tensors are defined with respect to some vector space, V. I sometimes call it the base space, but I don't know if there's a standard word for it. In differential geometry, V, happens to be a tangent space, that is, the space of all vectors tangent to a given manifold at some point. Its vectors are called tangent vectors, or sometimes contravariant vectors.

The type or valence of a tensor, denoted (p,q), specifies that it's a function of q vectors of V and p vectors of the dual space of V. The dual space is denoted by V*, and is a vector space over the same algebraic field as V, whose vectors are scalar-valued linear functions of exactly one vector of V. In the context of differential geometry, the dual space is called the cotangent space. Its vectors are called dual vectors, cotangent vectors, covectors, covariant vectors, linear functionals, linear forms or 1-forms. I think that the names 1-forms and cotangent vectors are used only when V is a tangent space to a manifold, and I think the other names are general. Dual vectors are tensors of valence (0,1).

The value p+q is sometimes called the order of the tensor, sometimes its rank, although the latter usage conflicts with the use of rank in linear algebra. Given a choice of basis for V, there's a natural way of defining a certain convenient basis for V* called a dual basis, and hence for higher order tensor spaces. I won't define it here, but all introductory books on the subject will.

V happens to be isomorphic to (V*)*, the dual space of the dual space. So if w is a dual vector, and v a vector of V, then we can define v(w) = w(v), and classify vectors of V as tensors of valence (1,0).

Scalars are defined as tensors of valence (0,0), being functions of no vectors of V or V*.

If the dimension of V is n, tensors of valence (p,q) for a particular choice of p and q, say for example (1,1), constitute the vectors of an n^(p+q) = n²-dimensional vector space, also called a tensor space. In this case, they can be represented in a particular basis as an nxn matrix of n² components. Thus, you can add together tensors of the same valence. Higher order tensors, i.e. those where p+q > 2, can be represented in a particular basis as higher dimensional arrays of components, but in the context of differential geometry all of these tensors are thought of as existing independently of any particular arbitrary choice of coordinate representation. They're said to be invariant, even if their components change, in the sense that they output the same real number, regardless of what coordinates may be used.

In the context of general relativity, people will talk about whether a certain quantity is or is not a tensor. They mean: "Is this quantity a tensor field, whose tensors are defined with respect to the tangent spaces at each point of the manifold we are using to model spacetime (and thus gives the same results when it acts on q tangent vectors and p cotangent vectors no matter what coordinates we use)?" But that's a bit of a mouthful!

Confusing GR jargon to watch out for: The word covariant has two meanings. In the name covariant derivative, it means invariant (in the sense described above). In the name covariant vector, a synonym for dual vector, it means a vector whose components, if you choose to represent it in component form, using a dual basis, will transform by the same rule as the basis you're using for the tangent space.

 

[arkajad]

In the context of general relativity, people will talk about whether a certain quantity is or is not a tensor.

In differential geometry there are first-order geometrical objects and second-order objects. Tensors are first-order objects - they are born from the tangent space. Connection coefficients (for instance Christoffel symbols) are second-order geometrical objects. This is discussed in monographs of Sternberg, Kobayashi, Sharpe, but not well known to ordinary relativists.